Ch. 13 - DNA

Question 1

Griffith’s experiment

Answer 1

In 1928, when Griffith inoculated mice with a mixture of living R-type (rough type - nonvirulent) bacterial cells and heat-killed S-type (smooth type - virulent) bacterial cells, the mice died of pneumonia; Griffith concluded that in the presence of dead S-type pneumococcus cells, some of the living R-type cells has been transformed into virulent S-cells (transforming factor = DNA)

Question 2

Avery, McLeod, McCarty experiment

Answer 2

In 1944, Avery et al conducted experiments with the Streptococcus pneumoniae bacteria where protein, RNA and DNA were eliminated from the heat-killed S-type cells; these samples were mixed with live R-type cells and when the cells were treated with DNase (breakdown of DNA), the transforming activity was lost and R-type cells were not transformed into S-type cells; concluded that DNA is the transforming factor

Question 3

Hershey-Chase experiment

Answer 3

Used bacteriophage T2 virus to determine whether DNA or protein is genetic material; viruses were labeled with radioactive sulfur and radioactive phosphorus and mixed with bacteria; the bacteria were isolated and found to contain radioactive phosphorus; determined that DNA transferred into the bacteria and was responsible for redirecting the genetic program of the bacterial cell

Question 4

Rosalind Franklin

Answer 4

Used X-ray diffraction to visualize crystallized DNA; her data suggested that DNA is structured as a double-stranded helix; 10 nucleotides per full turn; 3.4 nm in length; diameter of 2 nm suggested that the sugar-phosphate backbone of each DNA strand must be on the outside of the helix

Question 5

Chargaff’s rule

Answer 5

In any DNA sample, the amount of purines equal the amount of pyrimidines, the amount of adenine equal the amount of thymine (A=T) and the amount of guanine equals the amount of cytosine (G=C); with a DNA sample, you can identify the percentage of the nitrogenous bases in the sample (ie. 10% adenine = 10% thymine, 40% guanine = 40% cytosine)

Question 6

Watson and Crick’s model

Answer 6

In 1953, Francis Crick and James Watson used model building to solve the structure of DNA: their model had the nucleotide bases on the interior of the two strands, with a sugar-phosphate “backbone” on the outside, the two DNA strands ran in opposite directions or antiparallel and that bases were purine-pyrimidine pairs

Question 7

DNA’s four key features

Answer 7

1) DNA is a double-stranded helix 2) DNA is usually a right-handed helix 3) DNA is antiparallel 4) DNA has major and minor grooves (major grooves are exposed to allow DNA binding proteins to attach and replicate or transcribe)

Question 8

Antiparallel strands

Answer 8

DNA replicates and grows in a 5’ to 3’ direction; both strands run in opposite directions

Question 9

Components of DNA replication in a test tube

Answer 9

Deoxyribonucleoside triphosphates dATP, dCTP, dGTP, and dTTP (dNTPs are monomers of DNA); DNA template; DNA polymerase; salts and pH buffer

Question 10

Base Exposure in the Grooves

Answer 10

The surface of the A-T and the C-G base pairs chemically distinct, allowing other molecules such as proteins to recognize specific base pair sequences and bind to them; the binding of proteins to specific base pair sequences is the key to protein-DNA interactions necessary for replication and expression of genetic information in DNA

Question 11

Semi-conservative replication

Answer 11

Each parent strand serves as a template for a new strand; two new DNA molecules each have one old and one new strand

Question 12

Conservative replication

Answer 12

Two DNA molecules are formed, one parental strand and one new strand

Question 13

Dispersive replication

Answer 13

Fragments of the original DNA molecule serve as template for assembling two new molecules, each containing old and new parts, perhaps at random

Question 14

Meselson and Stahl Experiment

Answer 14

Experiment demonstrated that DNA replication is semiconservative; used density labeling to distinguish between parent strands of DNA and newly copied ones; DNA was extracted and after two replication cycles, two bands of DNA were seen, one of intermediate density and one of light density. This result is exactly what the semiconservative model predicts: half should be 15N-14N intermediate density DNA and half should be 14N-14N light density DNA

Question 15

Requirements for DNA replication

Answer 15

DNA template, dNTPs, helicase, topoisomerase, single strand binding proteins (SSBs), RNA primase, DNA polymerase III, RNase H, DNA polymerase I, DNA ligase

Question 16

Replisome

Answer 16

Complex molecular machine that carries out DNA replication; forms two new double stranded DNA sequences; remains stationary while DNA moves through the complex; includes a helicase, a polymerase and a sliding clamp; the replisome assembles at a region of the DNA, called the replication fork, where the double-stranded DNA is separated into two individual strands, which are both subsequently copied in the 5′ to 3′ direction of the DNA

Question 17

DNA replication

Answer 17

The pre-replication complex binds to the ori (origin point) and helicase starts to unzip the double-stranded helix forming a replication fork; single-strand binding proteins coat the separated strands of DNA near the replication fork, keeping them from coming back together into a double helix; primers are short starter strands that are synthesized by RNA primase and are placed at the start of the new strand; DNA polymerase III then adds nucleotides to the 3’ end of the primer until replication of that section of DNA is complete; RNAase H removes the RNA primer, DNA pol I replaces the RNA primer with DNA, DNA ligase catalyzes the formation of the final phosphodiester linkage of the backbone

Question 18

DNA polymerase structure

Answer 18

Large protein enzyme that is shaped like an open right hand with a palm, thumb and fingers; the palm is the active site of the enzyme that brings together the dNTPs with the DNA template; the finger region has precise shapes that recognize the different nucleotide bases (dNTPs)

Question 19

Topoisomerase

Answer 19

This enzyme prevents the DNA double helix ahead of the replication fork from getting too tightly wound as the DNA is opened up. It acts by making temporary nicks in the helix to release the tension, then sealing the nicks to avoid permanent damage

Question 20

Leading strand

Answer 20

During DNA replication, at the replication fork, the leading strand grows at the 3’ end as the fork opens

Question 21

Lagging strand

Answer 21

During DNA replication, the lagging strand is oriented so that as the fork opens up, its exposed 3’ end gets farther and farther away from the fork; this strand is made in fragments because, as the fork moves forward, the DNA polymerase (which is moving away from the fork) must come off and reattach on the newly exposed DNA

Question 22

Okazaki fragment

Answer 22

While the leading strand grows continuously forward, the lagging strand grows in shorter, “backward” stretches with gaps in between them; each fragment requires its own primer synthesized by RNA primase;

Question 23

Sliding DNA clamp

Answer 23

A protein that keeps the polymerase-DNA complex stabilized and in close contact; prevents DNA polymerase from dissociating from the DNA template; the clamp increases the efficiency of polymerization by keeping the enzyme (DNA pol) bound to the substrate

Question 24

Telomere

Answer 24

Eukaryote chromosomes have repetitive sequences (~2500 times) at the ends; on lagging strand, when the terminal Okazaki primer is removed, no DNA can be synthesized to replace it; at the end of the chromosome, however, there is no neighboring Okazaki fragment to provide the needed hydroxyl to start the DNA replacement, resulting in incomplete replication of the chromosome end; this short piece of DNA 
is removed; chromosome
becomes shorter w/each
replication; after many 
divisions, genes may be 
lost and cell dies

Question 25

Telomerase

Answer 25

Continuously dividing cells (ie. bone marrow stem cells) have telomerase, an enzyme that catalyzes addition of lost telomeres; typically expressed in most cancer cells which allow the cells to keep dividing

Question 26

DNA repair

Answer 26

1) Proofreading: as DNA polymerase adds a nucleotide to a growing stand, it recognizes mismatched pairs and replaces the base 2) Mismatch repair: newly replicated DNA is scanned for mistakes by other proteins and the protein complex corrects the mismatch 3) Excision repair: NER enzymes constantly scan DNA for damaged bases, which are excised and DNA pol I adds the replacement bases

Question 27

Xeroderma pigmentosum

Answer 27

Genetic disorder that lacks an excision repair mechanism that repairs damage caused by UV radiation; most commonly due to a defect in a NER gene; thymine dimers accumulate in skin cells which interferes with base pairing during replication